Find all real and imaginary solutions to the equation.
16, x, cubed, minus, 16, x, squared, minus, x, plus, 1, equals, 0
16x
3
−16x
2
−x+1=0

We are given the cubic equation:

\[
16x^3 - 16x^2 - x + 1 = 0
\]

To find all real and imaginary solutions, we'll solve it step by step:

### Step 1: Use Rational Root Theorem

The Rational Root Theorem suggests that potential rational roots can be factors of the constant term (1) divided by factors of the leading coefficient (16).

Possible rational roots are: \( \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}, \pm \frac{1}{16} \).

### Step 2: Test Possible Roots

We'll test these values using synthetic division or direct substitution to see if any are roots.

#### Test \( x = 1 \):

Substitute \( x = 1 \) into the equation:
\[
16(1)^3 - 16(1)^2 - (1) + 1 = 16 - 16 - 1 + 1 = 0
\]
Thus, \( x = 1 \) is a root.

### Step 3: Factor the Polynomial

Since \( x = 1 \) is a root, we can factor \( (x - 1) \) from the cubic equation. Perform synthetic division on \( 16x^3 - 16x^2 - x + 1 \) by \( x - 1 \).

\[
\begin{array}{r|rrrr}
1 & 16 & -16 & -1 & 1 \\
  &   & 16  & 0   & -1 \\
\hline
  & 16 & 0   & -1  & 0 \\
\end{array}
\]

The quotient is \( 16x^2 - 1 \), so the factored form of the cubic is:
\[
16x^3 - 16x^2 - x + 1 = (x - 1)(16x^2 - 1)
\]

### Step 4: Solve the Quadratic

Now, solve \( 16x^2 - 1 = 0 \):
\[
16x^2 = 1
\]
\[
x^2 = \frac{1}{16}
\]
\[
x = \pm \frac{1}{4}
\]

### Step 5: Final Solutions

We have the solutions \( x = 1 \), \( x = \frac{1}{4} \), and \( x = -\frac{1}{4} \).

Thus, the real solutions to the equation are:
\[
x = 1, \quad x = \frac{1}{4}, \quad x = -\frac{1}{4}
\]

Since all solutions are real, there are no imaginary solutions.

### Summary:

The real solutions are:
\[
x = 1, \quad x = \frac{1}{4}, \quad x = -\frac{1}{4}
\]

Would you like more details or have any questions?

Here are 5 related questions you could explore:
1. How can synthetic division be used to factor polynomials?
2. What is the Rational Root Theorem and how does it help in solving polynomials?
3. Can we solve the cubic equation graphically to verify the roots?
4. What are complex (imaginary) roots and when do they appear in cubic equations?
5. How do you handle cases when the cubic equation has no rational roots?

**Tip**: Always test small rational roots first when solving polynomials using the Rational Root Theorem, as they are often easier to check!

Find all real and imaginary solutions to the equation 16x^3 - 16x^2 - x + 1 = 0.

Learn how to solve the cubic equation 16x^3 - 16x^2 - x + 1 = 0 by applying the Rational Root Theorem and synthetic division. Follow a step-by-step guide to factor the polynomial and find the real solutions, suitable for high school students in Grades 10-12.

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